Question

The following reaction was allowed to come to equilibrium $A\left( g \right) + 2B\left( g \right) \rightleftharpoons C\left( g \right)$ . The initial amounts of reactants placed into a $5{\text{ }}lt$ vessel were $1$ mole of $A$ and $1.8$ mole of $B$ . After the reaction reached equilibrium. $1$ mole of B was found. Calculate the value of ${k_c}$ for this reaction.

Answer 1

In the given question firstly we have to define all the given details and the molar relations and the concentrations that are given in the problem statement. Now by that and the equation we get the relation through which we can get the value of mole $x$ by comparing it to the given data in the question that is $1$ mole of B was found.

Complete step by step answer:
Here according to the question statement we first have to take out the important details we get from the solution process:
i.The given reaction for the equilibrium process is : $A\left( g \right) + 2B\left( g \right) \rightleftharpoons C\left( g \right)$
ii.The amount of reactant at the start of the entire process : $5{\text{ }}lt$
iii.Given mole of the component $A$: $1$
iv.Given mole of the component $B$: $1.8$
Now the step by step answer to the problem is :
Step 1 : So here we have to use the basic unitary method and the basic analysis altogether in the first step.
$A\left( g \right) + 2B\left( g \right) \rightleftharpoons C\left( g \right)$
Initial and final mole of the $A,B,C$ are :
${\text{A B C}} \\\ {\text{1 1}}{\text{.8 0}} \\\ {\text{1 - x 1}}{\text{.8 - 2x x}} \\\$
Step 2: Now according to the question:
$1.8 - 2x = 1$
So we just have to solve and then we can get the answer of the values:
$1.8 - 2x = 1 \\\ 2x = 0.8 \\\ x = 0.4 \\\$
Step 3: So for the final step of the problem we need the moles per substance. So that is :
$[A] = \dfrac{{1 - x}}{5} = 0.12M \\\ [B] = \dfrac{1}{5} = 0.2M \\\ [C] = \dfrac{{0.4}}{5} = 0.08M \\\$
Step 4: Now the value for the ${k_c}$ should be by the formula as follows :
${k_c} = \dfrac{{0.08}}{{0.12 \times {{(0.2)}^2}}} \\\ {k_c} = 16.67 \\\$
Therefore we get the final answer for the given problem as ${k_c} = 16.67$ .

Note:
The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.